Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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+ | <math>\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} </math> | ||
==Solution== | ==Solution== |
Revision as of 15:46, 29 November 2015
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution
Triangle is isosceles, so . Using the symmetry of pentagon , notice that . Therefore, .
Since , . From this, we get .
Since , . Since , . Thus,
Therefore,
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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